[SOLVED] Does GCC have features to detect use-after-free bugs?

Issue

Here is my code snippet:

#include <stdlib.h>
#include <stdio.h>

typedef struct node
{
    char key;          // value
    struct node *next; // pointer to the next element
} Node;

int main(void)
{
    Node *n = malloc(sizeof(Node));
    n->key = 'K';
    n->next = NULL;

    free(n);

    printf("%c\n", n->key);
}

When the above snippet is compiled and run...

ganningxu@Gannings-Computer:~/test$ gcc test_faults.c -o test_faults; ./test_faults
K

ganningxu@Gannings-Computer:~/test$ clang test_faults.c -o test_faults; ./test_faults
K

There are no compiler warnings or errors when the freed memory is accessed. Is there any way to force the compiler to show such errors?

Solution

When compiling with GCC, you can use -fsanitize=address so that “Memory access instructions are instrumented to detect out-of-bounds and use-after-free bugs.” This will of course affect program performance, and it may also change the program behavior if there are bugs.

Answered By - Eric Postpischil